Lamp Testing

please do not post your Homework questions here

The lamp above your head when you finally figure it out for yourself!

Do you want to know which lamp glow brighter or which glow brightest?

hi

60w bulb give more light

since lamps have a high cold resistance and a low hot resistance, and aa high wattage lamp has lower resistance than a low wattage lamp and all current are the same in series you cannot reliably predict the end yemperature of each lamp or its resistence. In general the 60W will glow brightest and the 200W dimmer and the 100W in between.. Wil they make visible light? The 60 will be visible I estimate, but the 200W might not get hot enough. The 100W ??,

"since lamps have a high cold resistance and a low hot resistance"

Not unless they are carbon filaments. That's why we talk about inrush current on an incandescent lamp. Tungsten has a positive temperature coefficient (0.004403/K) as do most metals.

Yes, I knew that and spoke in error. Cold inrush is the correct situation and escribes tungsten filaments

My understanding is that bulbs have very low cold resistance; not high cold resistance. That is why if you happen to energize the bulb at the "Peak of the weave form it burns out at a brightness of a flash.

You have the resistance versus temperature coefficient correct. The hot resistance of a tungsten filament is about 10 times the cold resistance. When the bulb is energized with respect to the waveform makes no difference because the thermal time constant of the filament is many times longer than 1/2 cycle (at 60 Hz).

Don't you have these bulbs? Connect those in series and observe and inform the findings to this forum

Considering the lamps as purely resistive, and simply applying ohms law to the series circuit,first find individual lamp resistances which come out to be 264.5 ohms,529 ohms and 881 ohms(Apply Power=(Voltage)*(Voltage)/current).

Now that lamp grows the brightest to which theres maximum flow of current.But the flow of current is restricted by resistance.Hence, higher the resistance,lower the current and lower brightness.Hence you can infer easily that 200W lamp having least resistance will glow the brightest.

It seems to me that the wires connecting the bulbs to each other have the lowest resistance. Why wouldn't those wires glow btightest?

This is inefficient. Why not purchase a three phase transformer and bleed off two 115 volt lines and a common ground so to use 115 volt lighting instead.

Since there are threads with 60W and 200W advocating , I will take the average 130 W and vote for nearest - the 100W

In homework anyway best to put the tick on the middle one you know (professor puts the optimum in middle and the others on either side.

PS: I expect at least 10GA points on this brilliant logic and answer

I suggest one more way, which you will certainly agree:

Take a nationwide pole and find it out. The base with only CR4 subscribers is too small.

IFF the filament wires for all three are the same diameter of the same material, then they will all be equally "hot" and thus equally bright. (Just another opinion)

In this condition they would all give off different amounts of energy.

Since it is probable that the filaments are not identical material, then it would not be possible to predict where each filament was on its warming curve relative to the others.

By the way, what do you mean by "brighter"? Do you mean energy released, or do you mean the colour temperature of the filament or do you mean energy per unit area?

There is really insufficient definition of your question to provide a "correct" answer, only qualified guesses.

who is asking this question. An engineer, would be engineer or ... Why not do this siple practical and observe the findings.

I suggest that the bulb having the highest percentage of its recommended working voltage across it's terminals, would be the brightest. That would be the 60 watt bulb.

I was wondering how long it would take for someone to come up with a GA. Have most of us forgotten Ohm's Law?

Did I say Ohm's Law? Sorry. I meant to say Kirkoff's Law. Duh!

Regards.

No sir! it was a simple Ohms-Law question, not for "Kirkoff's Law" question.

I have to disagree with you. In order to determine which of the three lamps in a series circuit would have the highest voltage drop would require the use of Kirkoff's law.

Kirkoff's laws(rules) are only needed in a complex loop where the currents are NOT the same everywhere. This is a simple loop with one voltage source and three resistors. Ohm's law is all that is needed, together with the knowledge that the operating resistance goes up rapidly with temperature, and that temperature is dependent on the I²R losses in each bulb. With the same current flowing through all three lamps, that current will be a tiny fraction of the normal current for the 200W lamp, so it will barely heat at all, and its resistance will hardly change from its cold resistance. The 60W lamp will have the highest fraction of its normal current, so it will heat the most, which raises its resistance, increasing the fraction of the voltage 'allocated' to it. Thus another GA to welderman's post #16.

When dealing with potentials, Ohm's Law states: VOLTAGE = CURRENT * RESISTANCE. And, Kirkoff's Law states: THE SUM OF THE VOLTAGE DROPS AROUND A CIRCUIT WILL BE EQUIAL TO THE VOLTAGE DROP FOR THE ENTIRE CIRCUIT. Sound familiar?

Sort of...

I remember it as: "The sum of the voltage drops around a circuit will be zero." The circuit includes both the source(s) and the load(s).

A well described reply.

As you wish

Dear Sir.

I may add to your comment to elaborate:

And highest resistance will drop highest voltage drop across it in a Series circuit.

Regards

The 60W lamp as it has highest resistance & minimum current requirement.

Highest voltage drop across it X I [which is common for all lamps] =s highest power.

Note to remember:

lowest power-lamp has highest element resistance.

And highest resistance will drop highest voltage drop across it in a Series circuit

And highest resistance in a parallel circuit will pass minmum current through it. So minimum power dissipated across it as min current X equal Voltage across =s min power dissipation.

Regards

It does rather depend upon the voltage rating of each lamp, which hasn't been stated in the question!

You missed the last sentance. Each lamp is rated for 230V

Interesting question! In general, I "might" expect that the 60W lamp will be first to light up and glow brighter but I have not yet seen the following line of reasoning.

The filament can be a straight piece of tungsten wire or a coil, or a coiled coil. The difference between the 60 Watt filament and the 200 Watt filament will be enough that it will be quite easy to make general statements about them. As to the 60 Watt lamp and the 100 Watt lamp, there is some room for trickery.

Consider that the 60 Watt lamp filament is made of a given length L(60) of tungsten wire of diameter D(60). The 100 Watt lamp is assumed to be made of a little more of the same material with a length of L(60)*100/60 and diameter D(60). But what if the filament used was of length L(60) with a diameter of D(60)*60/100?

In the second case (100 W using a filament length same as the 60W) the mass of the coil and the rate of change of resistance come into play. Also remember that the initial current is the total voltage divided by the total initial resistance. As the coil responds to the current, there could be a race as to which coil increases its resistance the fastest. Then you can talk about Ohms law. Since none of the coils will have full voltage applied, they will not achieve the calculated operating resistance used in many of the explanations above. Without knowing more about the specific coil (or filament, to be exact) I would say that all bets are off. I've been tricked before by questions like this. There is still a lot of missing information about the three lamps being compared.

I wouldn't put it past some professor to try to blow his students minds with an observation that defied (assumed) logic.

You have a primary flaw in your logic. For a given supply voltage, (these were all given as 230V), a higher power lamp must have more current. I=P/E, I₆₀=60/230=0.26Amps, I₁₀₀=100/230=0.43Amps, and I₂₀₀=200/230=0.87Amps, when connected singly to 230V.

The operating (hot) resistance of the lamps must then be R=E/I, R₆₀=230/0.26=885Ω, R₁₀₀=230/.43=535Ω, and R₂₀₀=230/0.87=263Ω.

If all the filaments are tungsten of the same diameter, then their lengths will be directly proportional to their resistances. Let's assume that the 200W lamp has a filament 0.50" long. Then the 100W lamp would have to have a filament 0.50"*535/263=1.0" long (there is some rounding error), and the 60W lamp would have to have a filament 0.50"*885/263=1.7" long.

Note that the LOWest power lamp has the LONGest filament (highest resistance). In fact that is the reason that high voltage low power lamps have such a short lifetime - In order to get the required high resistance in a reasonable length of filament, the filament must be very thin. This means initially weak, and rapidly getting weaker as tungsten evaporates from the filament (and condenses on the glass, making the inside of the bulb darker)

Now the cold resistance of tungsten incandescent lamps is roughly one tenth of the hot resistance, so the initial current when first turned on is roughly 10 times the normal hot current. That high current makes the loops of the coil attract each other magnetically, which is why incandescent lamps usually burn out when first turned on, rather than after operating some time.

Actually, the flaw is in the assumption that the coils would be the same wire diameter or composition and different only in length. Consider that the temperature of the coil goes from ambient to nearly 1200ºC in order to incandesce. If 60 Watts of energy is displaced over 1.7 inches of tungsten filament and produces 1000 lumens, how is it possible that 200 Watts displaced over 0.5 inches is anywhere near the operating temperature that a given tungsten filament wire can tolerate?

Secondly, all your calculation need to be done at something close to the cold temperature values not the hot temperature values since all three are in series and therefore not able to reach the normal operating values of resistance. So, what seems to be a straight forward problem is not, in fact, as simple as it appears.

While this thread has nearly run its course, I think I'll begin a new one titled "Coil Testing" on which I have spent many hours studying. It is truly a "related rates" problem that I found to be somewhat difficult to predict.

As to the magnetic effect on the coil causing a problem on startup, I think that is altogether too easy of an assumption as to cause and effect. Burnout would be a lot more common if that effect was significant. Instead, I think you will find that there is a brief uneven heating caused by the heat sink effect of the clamp wire. Typically a coil heats up in the middle first and then spreads rapidly toward the clamps (connection to the outside world). The rapid temperature rise places significant mechanical stress on the filament wire and the coil expands proportionally with the rise in temperature.

Easier calculation is R = V²/ P

So

R₆₀= 230²/60 = 881.67

R₁₀₀ = 529

R₂₀₀ = 264.5

The total resistance of circuit = 1675.17

I = 0.137A

The power output by each is now I²R_n

ie P₆₀ = 16.62W

P₁₀₀ = 9.972 W

P₂₀₀ = 4.986W

This solution is assuming that your assignment submission date is over.

If you want the formula,

Total

R = V² (1/P₁+1/P₂+...)

Current = V/R = 1/(V*(1/P₁+1/P₂+...))

Power for each resistor

P_n = I²R_n = V²/P_n / {(V*(1/P₁+1/P₂+...))}²

= 1/P_n(1/P₁+1/P₂+...)²

eg

P60 = 1/60 *(1/60+1/100+1/200)² = 16.62W

Assuming - each load is purely resistive - else replace R with Z and the additions will be phasor addition

Each one is rated for same voltage else the equation becomes not so simple.

Except that, in the series circuit, the 200W lamp will be essentially cold, dropping its resistance to something on the order of 30 Ω. As NYOJ has indicated, the exact temperature of each filament will depend on lots of other factors, but the net effect of temperature is to reduce the energy lost in the highest power bulb, shifting proportionally more to the other two. The same thing happens between the other two: The larger one will be cooler than the smaller one, although warmer than the largest (highest power) filament. This will shift even more power to the smallest one.

I have done this experiment dozens of times as a classroom demonstration, with different brands and types of bulbs, and the result is always the same: The Largest bulb does not glow visibly, the middle one glows barely but visibly, and the smallest one glows at a little less than its normal brightness. Of course I have done this with 120V lamps, not 230V ones, but I am virtually certain that the results will be the same.